Exponential Equations

[Abdulla]

2^x - 2^(x-2) = 3(2^13)

What is the value of x?

A. 9
B. 11
C. 13
D. 15 (correct answer)
E 17

I knew that the trick is how to factor 2^x-2^(x-2) but can't find the answer. I want to understand how we reach 2^(x-2) 2^(2-1) ?

Thanks in advance

[Eric_J]

Abdulla-

the trick, as you surmised, is to factor the expression.

2^x - 2^(x-2) factors to:

2^(x-2) X [(2^2 - 1)] (because [2^(X-2)]X[2^2] = 2^x

so you have

[2^(x-2)] X [4-1] which equals

3*[2^(x-2)]

so, if we know that the expression above equals 3(2^13)

so, we can tell by inspection that x-2 = 13 which solves to x=15.

hope that helps,

Eric

[Abdulla]

Thanks Eric..
its clear for me now and yesterday i spent almost 4 hours in order just to find and understand this rule ..
I found that
a^b - a^b-1= a^b(1-b^-1) = a^b-1(b-1) could you explain this rule ?

[RonPurewal]

not only have we already addressed this particular problem, but this thread is in the wrong folder.

[RonPurewal]

Thanks Eric..
its clear for me now and yesterday i spent almost 4 hours in order just to find and understand this rule ..
I found that
a^b - a^b-1= a^b(1-b^-1) = a^b-1(b-1) could you explain this rule ?

sweet lord, that baby needs some parentheses and some spaces.

it appears that you're writing this:
a^b - a^(b-1) = (a^b)(1 - a^-1) = (a^(b-1))(a - 1)
note the corrections appearing in blue, which may go a long way toward explaining why you don't understand the rule: because, as currently written, it's simply wrong.

the middle one (in the corrected version) is a valid factorization in most cases, but factorizations of that sort (where you pull negative exponents essentially out of the ether) are vanishingly rare on the actual test.
in addition, that factorization doesn't work unless a is guaranteed to be nonzero. if there are no such guarantees, then you have to go with the right-hand factorization.

to understand the right-hand factorization, just think about examples.
here's one:
a^6 - a^5
= (a^5)(a^1 - a^0)
= (a^5)(a - 1).
now just replace the "6" with the more general "b" (whereupon all the "5"s become "b - 1"s), and you're in business.